# Comparison of acids to dissolve phosphates

I am currently reading this German paper on the dissolution of phosphates by acids and chelants.

The authors do a "fast-test" of the dissolving power of different acids which goes as follows:

For each test, increasing quantities (10-50ml) of 0.1 n acids are added to 100 mg of tricalcium phosphate and agitated for 1 hour. The residue is filtrated and weighed.

For 0.1 n hydrochloric acid 20ml suffice for complete dissolution of 100mg tricalcium phosphate. Citric acid [...] shows >90% dissolution with 40ml.

Not being a chemist I am having trouble interpreting this. What quantity is effectively being controlled by specifying 0.1 n acid? The paper is from the 60s. I thus suspect that n stands for normality but I am not sure.

What exactly is being measured?

1. The speed at which different acids dissolve the material? (Why then vary the volume of acid?)
2. The quantity of acid needed to dissolve a given of material? (Measured in what exactly? What does the volume effectively stand for? Moles?)
3. Something else?

## Edit 1

J M's answer makes this much clearer. I still have some problems though:

1. Regarding normality: Does this ensure that the same concentration of protons is in each solution? I assume this is not the same as "constant PH" as some acids may be strong acids while others may be weak?

2. If even more citric acid was added, there would still remain 10% residue? Why did they not try with 50ml citric acid? With this test there are two variables that change: volume and residue. Does this not make it difficult to compare the results of different acids? I would understand a test like this: how much acid is needed to dissolve the material 100%. Then one could compare easily.

3. Regarding weak acids: Once one of the available protons from the citric acid has reacted, should not another previously non-dissociated acid molecule dissociate (to keep the dissociation proportion in equilibrium) and keep everything going to dissolve the material without any residue? Given infinite time, why do not 30ml citric acid also dissolve 10 mg of 100 mg of tricalcium phosphate?

4. I am still not quite sure however what the point of this test is. Should it not be reasonably easy to calculate this using the reaction equations and $\mathrm{p}K_\mathrm{a}$ of the acids? (See reaction equation below.)

5. Thought experiment: If not the acid quantity was the limiting factor but the amount of solution what would limit the reaction: i.e. infinite supply of acid molecules (high acid concentration) and infinite supply of phosphates but finite volume for the solution to take up the reaction product: Would the reaction be limited by the solubility of the reaction products? I assume, in this case a strong acid would be advantageous, right?

Reaction Equation

If I understand the papers I read correctly, one can dissolve 15 moles of $\ce{Ca_3(PO_4)_2}$ with 60 moles of $\ce{HCl}$ or 20 moles of citric acid (disregarding the fact that it is a weak acid?!):

$$\ce{ 15 Ca_3(PO_4)_2 + 60HCl -> 15Ca(H_2PO_4)_2 + 30 CaCl_2 }$$ $$\ce{ 15 Ca_3(PO_4)_2 + 20 CitH_3 -> 10 Ca(Cit)_2 + 25 Ca(H_2PO_4)_2}$$

N, or Normality is a way of standardizing the # of acid equivalencies a particular acid has across multiple acids. Acids that have only 1 acidic hydrogen for example $\text{HCl}$, the Normality will equal Molarity.

Normality is defined by:

$$Normality = \frac{c_i}{f_{eq}}$$

Where $c_i$ is the molar concentration of the acid, and $f_{eq}$ is the equivalency factor. In the cases of something like $\text{H}_2\text{SO}_4$, the equivalency factor is 0.5 because it is a diprotic (two proton) acid.

This test also is showing the effect of weak vs strong acids. $\text{HCl}$ is a strong acid, so it dissociates essentially 100%, while citric is a weak acid. A weak acid exists in equilibrium between the dissociated acid species, and the neutral species. This equilibrium is defined by the acid-dissociation constant $K_a$, or it's logarithmic version $pK_a$.

For citric acid though, it is triprotic, meaning there are 3 acidic hydrogens, so it's $f_{eq}$ value is $1/3=0.3\overline{3}$

1) The speed is being measured to a certain point. However, the chance is that all of these reactions are complete at this point. The hour timepoint is to normalize everything to the same time frame. But as indicated, the reactions are likely complete at this point. What is really being measured is your second point.

2) Absolutely, the quantity of acid needed to dissolve the material is being tested. As you use more and more volume of acid, you are providing larger amounts of acid to the reaction. This helps it dissolve more of the powder (as there is more acid now present), and also increase the rate of the reaction.

They standardize the reaction to normality because if they used molarity, 1 mole of citric acid has 3 acid equivalents, while $\text{HCl}$ has only 1 acid equivalent.

• Thanks J M. This really helped me along the path of understanding. However, as some things became clearer, new questions came up. Would you mind taking a look at the addition to my question? – ARF Aug 14 '12 at 23:31
• Sure I will do my best to take a look later this evening and add an answer to those other points. One difficulty is that I do not read German, so I cannot read through the paper to digest more information. – J M Aug 15 '12 at 0:54
• Thanks. The paper is from an age where German was still acceptable as publishing language. - Things have changed quite a bit and it is obvious why... – ARF Aug 15 '12 at 9:42
• @ArikRaffaelFunke Just wanted to say I will be posting a response tonight or tomorrow morning, just finished submitting my MS Thesis last night – J M Aug 19 '12 at 14:51
• There really is no rush at all. Take all the time you want. I am merely curious because as a physicist I never even got an introduction into the most basic concepts in chemistry. Thanks for allowing me to pick your brain. - BTW: Congratulations on submitting your MS thesis. That is quite an accomplishment! – ARF Aug 19 '12 at 18:35

Indeed, n (pretty weird, chemists would use the capital N) is used for normality.

In their fast test, the authors try to dissolve 100 mg (0.322 mmol) of $\ce{Ca3(PO4)2}$ with different acids and buffers at room temperature while shaking for 1 hr.

For comparison, they state that 20 mL of 0.1N HCl is sufficient to dissolve all the triphosphate.

In the case of the other solvents (solution according to STUBY, etc.), where solution was incomplete, they determine the weight of the dried residue (i.e. how much solid material was left) and try to correlate it with the volume of the solvent used (cf. Fig. 1). From the volume used and the concentration of the citric acid in the stock solution the absolute amount can be calculated.